$\displaystyle
\ln y \;=\;11\bigg[\frac{x^2}{2} - \frac{11}{2}\ln(x^2+11)\bigg] + C
$
can someone rewrite this in terms of y?
I know you need to multiply by e, but I don't know what to do from there.
Thanks.
Hello,
It's not "multiplying", but composing with the exponential function.
$\displaystyle \exp(x)=e^x$
$\displaystyle \frac{1}{11} ~ \ln y=\frac{x^2}{2}-\frac{11}{2} ~ \ln(x^2+11)+\frac{C}{11}$
$\displaystyle \ln \left(y^{1/11}\right)=\frac{x^2}{2}-\ln \left(\left(x^2+11\right)^{11/2}\right)+\frac{C}{11}$
$\displaystyle y^{1/11}=\exp \left(\frac{x^2}{2}-\ln \left(\left(x^2+11\right)^{11/2}\right)+\frac{C}{11}\right)$
$\displaystyle y^{1/11}=\frac{\exp \left(\frac{x^2}{2}+\frac{C}{11}\right)}{\exp\left (\ln \left(\left(x^2+11\right)^{11/2}\right)\right)}$
$\displaystyle y^{1/11}=\frac{\exp \left(\frac{x^2}{2}\right) \cdot \exp \left(\frac{C}{11}\right)}{(x^2+11)^{11/2}}$
And honestly, I don't think it is possible to write this in terms of y
It does look like there not too much you can do to simplify this, except the piece with ln.
$\displaystyle \ln y \;=\;11\bigg[\frac{x^2}{2} - \frac{11}{2}\ln(x^2+11)\bigg] + C
$
$\displaystyle y = e^{\frac{11}{2}x^2+C}*e^{\frac{-121}{2}\ln (x^2+11)}$
$\displaystyle y = e^{\frac{11}{2}x^2+C}*e^{\ln (x^2+11)^\frac{-121}{2}}$
$\displaystyle y =\bigg[(x^2+11)^\frac{-121}{2}\bigg]*e^{\frac{11}{2}x^2+C}$